Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.io.*;

 
public class Main {
 
	public static void main(String[] args) throws IOException {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(bf.readLine());
		double [] array = new double[n+1];
		array[0] =1;
		String[] line2 = bf.readLine().split(" ");
		for( int i = 0; i < n; i++) {
			Double k = Double.parseDouble(line2[i]);
			for(int j = i+1 ;j >= 0; j--) {
				if( j == 0)
					array[j] = array[j]* (1-k);
				else
					array[j] = array[j-1]* k + array[j]* ((double)1-k);
			
			}
		}
		double fin = 0;
		for(int i = n/2 + 1; i <= n; i++)
			fin += array[i];
		System.out.println(fin);
	       
	}

}